Wooden Sticks

发布时间：2014-10-22 19:01:33来源：分享查询网

E - Wooden Sticks
Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d
& %I64u
SubmitStatus
Description
There is a pile of n wooden sticks. The length and weight of each stick are known in advance. The sticks are to be processed by a woodworking machine in one by one fashion. It needs some time, called setup time, for the machine to
prepare processing a stick. The setup times are associated with cleaning operations and changing tools and shapes in the machine. The setup times of the woodworking machine are given as follows:
(a) The setup time for the first wooden stick is 1 minute.
(b) Right after processing a stick of length l and weight w , the machine will need no setup time for a stick of length l' and weight w' if l<=l' and w<=w'. Otherwise, it will need 1 minute for setup.
You are to find the minimum setup time to process a given pile of n wooden sticks. For example, if you have five sticks whose pairs of length and weight are (4,9), (5,2), (2,1), (3,5), and (1,4), then the minimum setup time should be 2 minutes since there is
a sequence of pairs (1,4), (3,5), (4,9), (2,1), (5,2).
Input
The input consists of T test cases. The number of test cases (T) is given in the first line of the input file. Each test case consists of two lines: The first line has an integer n , 1<=n<=5000, that represents the number of wooden
sticks in the test case, and the second line contains n 2 positive integers l1, w1, l2, w2, ..., ln, wn, each of magnitude at most 10000 , where li and wi are the length and weight of the i th wooden stick, respectively. The 2n integers are delimited by one
or more spaces.
Output
The output should contain the minimum setup time in minutes, one per line.
Sample Input
3
5
4 9 5 2 2 1 3 5 1 4
3
2 2 1 1 2 2
3
1 3 2 2 3 1
Sample Output
2
1
3
#include<stdio.h>
#include<string.h>
#include<iostream>
#include<algorithm>
using namespace std;
int main()
{
int n,m,l[5005],w[5005], t[5005],u[5005],d; int e[5005],r=0;
scanf("%d",&m);
while(m--)
{
memset(t,0,sizeof(t));
memset(u,0,sizeof(u));
memset(l,0,sizeof(l));
memset(w,0,sizeof(w));
scanf("%d",&n);
for(int b=0;b<n;b++)
scanf("%d%d",&l[b],&w[b]);//cin>>l[b]>>w[b];
for(int j=0;j<n-1;j++)
for(int k=j+1;k<n;k++)
{
if(l[j]>l[k])
{
d=l[j];l[j]=l[k];l[k]=d;
d=w[j];w[j]=w[k];w[k]=d;
}
else if(l[j]==l[k])
{
if(w[j]>w[k])
{
d=w[j];w[j]=w[k];w[k]=d;
}
}
}
int count,x=0 ;
for(int i=0;i<n;i++)
{
for(int j=0;j<=x;j++)
{ count =0;
if(t[j]<=l[i]&&u[j]<=w[i])
{t[j]=l[i];u[j]=w[i];count=1;break;}
}
if(count ==0)
{
x++;t[x]=l[i];u[x]=w[i];
}
}
if(n==0) x=-1;
e[r]=x+1;
r++;
}
for(int i=0;i<r;i++)
printf("%d\n",e[i]);
return 0;
}